Question

(3) (30 points) In Example 10.2, the moment of inertia tensor for a uniform solid cube of mass Mand side a is calculated for rotation about a corner of the cube. It also worked out the angular momentum of the cube when rotated about the x-axis - see Equation 10.51. (a) Find the total kinetic energy of the cube when rotated about the x-axis. (b) Example 10.4 finds the principal axes of this cube. It shows that one of the principal axis is in the direction ê=(1,1,1) which is along the body diagonal of the cube. The other two corresponding to the degenerate roots are any two mutually perpendicular vectors in the plane perpendicular to êŋ. So let us choose them as ês = 75 (1,-1,0) and és --a (1,1, -2). Show that if o is along ê,, that the angular momentum is in the same direction as ê, .. (c) If w is along the x-axis (i.e. O=(0,0,0) - see Equation 10.50), express this as a linear combination of ê, ê, and ê (d) Using the Euler's equations (see Equation 10.88), calculate the torque required to rotate a solid cube about one of its edges.

Heres example 10.2

Answer 1

Ja Hind! - Kolum lu of a cube roatating. we have given that . Moment of inertia od tensor I=83ma2 Ema²-ama2 T - na [ hma? Ima 2 zma? 7 er 8 -3 -37 2 m2 m2 13 M92/ 12 L-3 -3 8J @ when to cube is Roteating about ax-anis. Kine tie Energy (k.f.) = {I62 K.E. Iw? ja mal -3 § -3 11 : 1-3 -3 JT KEE marco" (3:35) Scanned with CamScanner